Problem: $ C = \left[\begin{array}{rr}-2 & 5 \\ 5 & 3\end{array}\right]$ $ A = \left[\begin{array}{rr}2 & -2 \\ 1 & 3\end{array}\right]$ What is $ C A$ ?
Because $ C$ has dimensions $(2\times2)$ and $ A$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ C A = \left[\begin{array}{rr}{-2} & {5} \\ {5} & {3}\end{array}\right] \left[\begin{array}{rr}{2} & \color{#DF0030}{-2} \\ {1} & \color{#DF0030}{3}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ C$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ C$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ C$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-2}\cdot{2}+{5}\cdot{1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ C$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{2}+{5}\cdot{1} & ? \\ {5}\cdot{2}+{3}\cdot{1} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ C$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{2}+{5}\cdot{1} & {-2}\cdot\color{#DF0030}{-2}+{5}\cdot\color{#DF0030}{3} \\ {5}\cdot{2}+{3}\cdot{1} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-2}\cdot{2}+{5}\cdot{1} & {-2}\cdot\color{#DF0030}{-2}+{5}\cdot\color{#DF0030}{3} \\ {5}\cdot{2}+{3}\cdot{1} & {5}\cdot\color{#DF0030}{-2}+{3}\cdot\color{#DF0030}{3}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}1 & 19 \\ 13 & -1\end{array}\right] $